Engine Thermodynamics Engine Testing and Instrumentation 1 Thermodynamics – study of heat related to matter in motion. Engineering thermodynamics mainly concerned with work producing or utilising machines such as engines, turbines and compressors together with the working substances used in the machines. Working substance – fluids: capable of deformation, energy transfer, common: air and steam Engine Testing and Instrumentation 2 Pressure F P = A 2 Force per unit area. Unit: Pascal (Pa) N / m 5 1 bar = 10 Pa 1 standard atmosphere = 1.01325 bar 1 bar =14.504 psi (pound force / square inch, 1N= 0.2248 pound force) 1 psi= 6894.76 Pa Engine Testing and Instrumentation 3 Psia vs Psig a- absolute g- gauge psia = psig + 1 atmosphere pressure e.g: 25 psig = 25+14.504= 39.504 psia 25 psig = 25 × 6894.76 Pa = 172368 Pa (gauge) = 172368 Pa +101325 Pa(1 atmosphere) = 273693 Pa (absolute) Engine Testing and Instrumentation 4 Phase Nature of substance. Matter can exists in three phases: solid, liquid and gas Cycle If a substance undergoes a series of processes and return to its original state, then it is said to have been taken through a cycle. Engine Testing and Instrumentation Process A substance is undergone a process if the state is changed by operation having been carried out on it Constant temperature process Isothermal process Constant pressure process Isobaric process Constant volume process Isometric process or isochoric process Adiabatic Process An adiabatic process is a process in which no heat is transferred. This is the case if the process happens so quickly that there is no time to transfer heat, or the system is very well insulated from its surroundings. Polytropic process A process which occurs with an interchange of both heat and work between the system and its surroundings. Nonadiabatic expansion or compression of a fluid is an example of a polytropic process. Engine Testing and Instrumentation 6 Energy Capacity of doing work Work A force is moved through a distance e.g. In a piston, work done = PA × L = P × AL = P()V2 − V1 Unit of work: Nm = J (joule) Power Rate of doing work P J/s = Watt L Engine Testing and Instrumentation 7 In general case, work V work done = 2 P dV ∫V 1 P P1 P2 V1 V2 V Engine Testing and Instrumentation 8 Polytropic Process PV n = C - constant – a law for the general case of expansion or compression n- the index of expansion or compression, or an polytropic exponent. E.g. air n =1.4 This is the general case of process for gas. Engine Testing and Instrumentation 9 Work done in polytropic process workdone V = 2 PdV ∫V 1 V =C 2 V −n dV ∫V 1 C −n+1 −n+1 = ()V2 −V1 −n+1 PV −PV = 1 1 2 2 n−1 Engine Testing and Instrumentation 10 Heat Temperature t (Celsius) = T-273.15 (Kelvin) Q – heat energy joules/kg Specific heat capacity: heat transfer per unit temperature: dQ c = dt Unit: joules/kg K (joules per kg per K) Calorific value: the heat liberated by burning unit mass or volume of a fuel. e.g. petrol: 43MJ/kg Engine Testing and Instrumentation 11 Enthalpy H = U + P V U – internal energy P – pressure v – volume Specific Enthalpy h =U/m= u + P v Engine Testing and Instrumentation 12 Principle of the thermodynamic engine Q Q-W Source Engine Sink W Engine Testing and Instrumentation 13 Thermal efficiency Work done W η = = Heat received Q Heat Engines An engine in which transfers energy results from difference in temperature. Engine Testing and Instrumentation 14 Mechanical Power Work done Power = Time Taken Unit: J/s= watt Electrical Power W = I V Unit: J/s=Watt Engine Testing and Instrumentation 15 The Conservation of Energy For a system Initial Energy + Energy Entering = Final Energy +Energy Leaving Potential energy = gZ 1 Kinetic energy = mC 2 2 Engine Testing and Instrumentation 16 Laws of Thermodynamics The zeroth Law If body A and B are in thermal equilibrium, and A and C are in thermal equilibrium, then, B and C must in thermal equilibrium. The first law W=Q Means if some work W is converted to heat Q or some heat Q is converted to work W, W=Q. It does not mean all work can convert to heat in a particular process. Engine Testing and Instrumentation 17 The second law Nature heat transfer will occur down a temperature gradient. The third law At the absolute zero of temperature the entropy of a perfect crystal of a substance is zero. Engine Testing and Instrumentation 18 Boyle’s law For perfect gas PV=C -constant Temperature remains constant. or P1V1 = P2V2 - constant T (Isothermal process) Charle’s Law V = constant T - constant P (Isobaric process) Gay-Lussac's law P = constant (Isometric process) T -- constant V Engine Testing and Instrumentation 19 Combined gas law PV = a (constant) T or PV P V 1 1 = 2 2 T1 T2 Let mR=a, where m is the mass, R - specific gas constant (air: R=287 J/kgK) PV = mR T - characteristic equation of gas Engine Testing and Instrumentation 20 Joule’s law Internal energy of gas is the function of temperature only and independent of changes in volume and pressure. The specific heat capacity at constant volume cv U 2 −U1 = mcv ()T2 − T1 (change in internal energy) The specific heat capacity at constant pressure c p U 2 −U1 + P(V2 −V1) = mc p ()T2 − T1 Engine Testing and Instrumentation 21 Polytropic Process n PV = C From n n P1V1 = P2V2 n −n −1/ n P1 V2 V1 V1 P1 = = and = P2 V1 V2 V2 P2 Engine Testing and Instrumentation 22 From the characteristice equation PV P V 1 1 = 2 2 T1 T2 n n−1 T1 P1V1 V2 V1 V2 = = = T2 P2V2 V1 V2 V1 and n−1 −1/ n n T1 P1V1 P1 P1 P1 = = = T2 P2V2 P2 P2 P2 i.e. n−1 T1 V2 = T2 V1 n−1 n T1 P1 = T2 P2 Engine Testing and Instrumentation 23 Entropy T (absolute temperature) s(entropy) s1 s2 Revisable heat transfer: Qrev dQ = T ds rev s2 Qrev = T ds ∫s 1 An isolated system can only change to states of equal or greater entropy. Engine Testing and Instrumentation 24 Entropy for gas Polytropic process n PV = C Heat transferred c − n cp dQ = Pdv c = c −1 cv dQ c − n P c − n R ds = = dv = dv T c −1 T c −1 v (combined characteristic eq.) where s – specific entropy v is a unit of V. s2 c − n v2 dv ds = R ∫s1 c −1 ∫v1 v c − n v2 s2 − s1 = R ln c −1 v1 Engine Testing and Instrumentation 25 c − n v2 s2 − s1 = R ln c −1 v1 c − n T1 s2 − s1 = cv ln n −1 T2 c − n P1 s2 − s1 = cv ln n P2 Engine Testing and Instrumentation 26 c p − n c − n T1 cv T1 c p − cvn T1 s2 − s1 = cv ln = cv ln = ln n −1 T2 n −1 T2 n −1 T2 − c (n −1) + c n − c n T = p p v ln 1 n −1 T2 T1 n T1 = −c p ln + (c p − cv ) ln T2 n −1 T2 n−1 T T n = c ln 2 − R ln 1 (c − c = R) p p v T1 T2 i.e T2 P2 s2 − s1 = c p ln − R ln T1 P1 Engine Testing and Instrumentation 27 T2 P2 s2 − s1 = c p ln − R ln T1 P1 Let s1 = 0, choose (arbitrarily) o T1 = 273.15K, (0 C) 2 P1 = 0.101MN / m = 1.01bar (1 standard atmosphere T P s = c ln 2 − R ln 2 2 p 273.15 0.101 Engine Testing and Instrumentation 28 Engine Testing and Instrumentation 29 Example: The air at T=288K and standard atmosphere is compress to P=5 times of standard atmosphere pressure with a temperature 456.1K. Calculate the entropy values. For air Cp=1005 J/kgK, R=287 J/kgK. Before compression 288 0.101 s = 1005 ln − 287 ln = 53 2 273.15 0.101 After compression 456.1 5* 0.101 s = 1005 ln − 287 ln = 53 2 273.15 0.101 No change in entropy. Engine Testing and Instrumentation 30 Isentropic Process An isentropic process is one during which the entropy of working fluid remains constant. T s Engine Testing and Instrumentation 31 Petrol Engine The cycle in a four-stroke cycle modern petrol engine is called the Otto cycle after German engineer Nikolaus Otto, who introduced it in 1876. It improved on earlier engine cycles by compressing the fuel mixture before it was ignited. Engine Testing and Instrumentation 32 (Courtesy of Tiscali ENCYCLOPAEDIA) Engine Testing and Instrumentation 33 Diesel Engine Fuel is injected on the power stroke into hot compressed air at the top of the cylinder, where it ignites spontaneously. The four stages are exactly the same as those of the four-stroke or Otto cycle. Engine Testing and Instrumentation 34 Mr Diesel’s first engine Engine Testing and Instrumentation 35 Engine Testing and Instrumentation 36 Nicolaus August Otto Four Stroke Engine 1876 Engine Testing and Instrumentation 37 Isentropic Compression or expansion T - temperature (K) p - pressure v - volume s - Entropy c p = specific heat at constant pressure cv = specific heat at constant volume gas constant R = c p − cv γ = c p / cv R 1 or =1− cp γ Engine Testing and Instrumentation 38 T2 p2 s2 − s1 = c p ln − R ln T1 p1 For reversible Compression and expansion, s2 = s1 Therefore T2 R p2 c p ln = ln T1 c p p1 Effect of pressure on temperature 1 1− γ T2 p2 = T1 p1 Equation of state: pv = RT Effect of volume of pressure γ p2 v2 = p1 v1 Engine Testing and Instrumentation 39 Engine Testing and Instrumentation 40 Otto cycle Otto Thermodynamic Cycle is used in all internal combustion engines.